To prepare an aqueous solution at 30 C, you have 80 grams of CaF2 with 670 grams

Question

To prepare an aqueous solution at 30 C, you have 80 grams of CaF2 with 670 grams of H2O, the density of the dissolution is 1.09 g/ml. The pressure of water vapor at 30 C is 26.90 mmHg. Calculate: a) molality b)freezing point of the disolution c) boiling point of the disolution d) vapor pressure of the dissolution e) mass percentage of the solution f) osmotic pressure To prepare an aqueous solution at 30 C, you have 80 grams of CaF2 with 670 grams of H2O, the density of the dissolution is 1.09 g/ml. The pressure of water vapor at 30 C is 26.90 mmHg. Calculate: a) molality b)freezing point of the disolution c) boiling point of the disolution d) vapor pressure of the dissolution e) mass percentage of the solution f) osmotic pressure

Explanation / Answer

(a)

Mass of solute, CaF2 = 80 g.

Molar mass of, CaF2 = 78 g/mol

Number of mole sof solute = mass / molar mass = 80 / 78 = 1.026 mol

Mass of solvent, H2O = 670 g. = 0.670 kg.

Now,

Molality = number of moles of solute / mass of solvent in kg

m = 1.026 / 0.670

m = 1.53m

(b)

Freezing point constant of H2O = 1.86 K.kg./mol

Freezing point of water = 0 degree C
NOw,

freezinf point depression = Kf * m

0 - Tf = 1.86 * 1.53

Tf = - 2.846 degree C

(c)

Boling point elevation constant of water, Kb = 0.512

Boiling point of water = 100 degree C
Now,

Elevation in boiling point = Kb * m

Tb - 100 = 0.512 * 1.53

Tb = 100.783 degree C
(d)

Number of moles of water = 670 / 18 = 37.22 mol

Mole fraction of solute = 1.026 / (1.026 + 37.22) = 0.02683

Vapour pressure of water at 30 degree C = 26.90 mm Hg

Now,

Relative lowering of vopur pressure = mole fraction of solute

(26.90 - p) / 26.90 = 0.02683

26.90 - p = 0.7217

p = 26.18 mmHg

(e)

Mass % of solute = 80 / (670 + 80 ) * 100 = 10.67 %

Mass % of solvent = 670 / (670 + 80) * 100 = 89.33 %

(f) Volume of 1.09 g. of solution = 1 mL

then, volume of 750 g. of solution = 1 * 750 / 1.09 = 688.07 mL = 0.68807 L

Molarity = number of moles of solute / volume of solution in L = 1.026 / 0.68807 = 1.49 M

Now,

Osmotic pressure = C R T = 1.49 * 0.0821 * 303.15 = 37.08 atm

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